Can you help me solve this system of equation using elimination.... please???

Question

i know how to use substitution but not elimination can you help please?


5x-4y=14
3y+3x=3

Answer 1

Multiply the equations so that when you summate them you have a variable eliminated so...

To eliminate y (one is -ve and the other is +ve)

Multiply the 1st eqn by 3
Multiply the 2nd eqn by 4

15x-12y=42
12y+12x=12

Sum the equations and you have:

27x=54
x=2

Substitute that in any equation to get y=-1

Subsitution for the whole thing would also equal:

x= 2
y= -1

Answer 2

5x - 4y = 14
3y + 3x = 3

5x - 4y = 14
3x + 3y = 3

Re arrange the second equation from 3y + 3x = 3 to 3x + 3y = 3

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Elimination of y. Solving for x

Multiply lthe top equation by 3 and the bottom equation by 4.

3(5x) - 3(4y) = 3(14)
4(3x) + 4(3y) = 4(3)

15x - 12y = 42
12x + 12y = 12
- - - - - - - - - - - - - - -
27x = 54

27x/27 = 54/27

Dividing the equation by 27

x = 2

Insert the x value into the equation

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Solving for y

5x - 4y = 14

5(2) - 4y = 14

10 - 4y = 14

-10 -10

Subtracting -10 from both sides


- 4y = 4

-4y/-4 = 4/-4

Dividing the equation by - 4

y = -1

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

5x - 4y = 14

5(2) - 4(- 1) = 14

10 + 4 = 14

14 = 14

- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - -

3x + 3y = 3

3(2) + 3(- 1) = 3

6 - 3 = 3

3 = 3

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The answer is

x = 2

y = - 1

Answer 3

To use elimination in a common system of linear equations, you want to get a common variable (x or y) to have a common coefficient (the number in front of x and y; in this case 5 and 3 for x and -4 and 3 for y).

You do this by finding least common multiple (although really, any common multiple will do) for two particlar coefficients. In order for this to happen, you need to manipulate each equation by multiplying it with the goal of changing one coefficient to a particular value.

In this case, let us look at the x variable. The coefficients are 5 and 3. The least common multiple of 5 and 3 is 15, and thus, 15 is the number we want to be in front of both x's.

So let us label 5x - 4y=14 as equation A
And let us label 3y + 3x = 3 as equation B

In order to make the 5 coefficient 15, we need to multiply equation A by 3 and equation B by 5. This gives us:

15x - 12y = 42 [A']
15x + 15y = 15 [B']

Notice that I gave the equations new names, A-prime and B-prime respectively. Also, I rearranged x and y in equation B'. This is allowed by the commutative law of addition.

At this point, the equations are such that we can "eliminate" by subtracting one from another. Since it is more complicated and error-prone to subtract negatives, we will subtract [A'] - [B']. This gives us:

0x - 27y = 27

At this point, simple algebra will arrive at the solution y = -1. After this, we can easily substitute y into one of the original equations to find x. Let us take equation B as an example.

3(-1) + 3x = 3
-3 + 3x = 3
3x = 6
x = 2

Therefore the solution by elimination is (2,-1).

A note is that some people prefer, instead of getting a common coefficient, getting coefficients that are additive inverses (such as 15 and -15). This allows the addition of two parts of a system instead of subtraction in the elimination phase. Either way, you will arrive at the final desired solution, and the only step left is to check. Hope this helps.. Good luck with your math studies!

Answer 4

Okay, rearrange the second one to start with x.

5x - 4y = 14
3x + 3y = 3

Now, draw a line underneath the two, like this.

5x - 4y = 14
3x + 3y = 3
---------------

Now, you know that when you add, you use a set-up similar to this, right?
Ex.
2
+2
----
4

Pretend that it's like that.

5x - 4y = 14
3x + 3y = 3
---------------

You are trying to "ELIMINATE" one variable by adding or subtracting. That means you need to get one variable in each equation to be the same, (i.e. -15y, 15y) with opposite signs. In this one, you should try and get y to have the same numbers in front because you already have opposite signs.

What is 4 x 3? Right, 12. You want the y's to be -12y and 12y. To do that, you need to multiply the entire first equation by 3 because 3 x -4 = -12.

That means you need to multiply the entire bottom equation by 4 because 3 x 4 = 12.

3(5x - 4y = 14)
4(3x + 3y = 3)
------------------

Now you get...

15x - 12y = 42
12x + 12y = 12
-------------------

Now, treat this like an addition problem, adding each column down.

15x - 12y = 42
12x + 12y = 12
-------------------
27x = 54

See? The y's are eliminated because -12y + 12y = 0

Now, solve for x.

27x = 54
x = 54/27
x = 2

Now, simply plug in the x value of 2 into one of the first equations. You should do it in the second one because it will be easier.

3(2) + 3y = 3
6 + 3y = 3
3y = -3
y = -1

There you go. Now you know that

x = 2
y = -1

Answer 5

Divide the second equation by 3, so you get x+y=1

Now you have two equations:
5x-4y=14 .............. (1)
x+y= 1 .................. (2)

Multiply the second by 4:
4x+4y=4 ................(3)

Adding (1) and (3) eliminate y to get
9x = 18, i.e. x = 2.

Now you can substitute x=2 in any equation to get y = -1

If you want to solve for y using elimination, you can multiply equation (2) by 5 and subtract the resultant from (1)

Answer 6

OK, the elimination method will attempt to "eliminate" one of the variables.

re-writting the equations:

5x - 4y = 14 (1)
3x + 3y = 3 (2)

Multiplying the first equation by 3 and the second by 4

15x - 12y = 42
12x +12y = 12

Adding the two equations:

27x = 54

Solving for x

x = 54/27 = 2

And, by substitution,

3y + 3(2) = 3
3y = -3

y = -1

Good luck.

Answer 7

i'm going to do the final 2, somebody else will probable do the 1st 2 for you: 3) Get each and every thing interior the 1st equation in terms of x: (13 - 5y)/2 = x Now placed all that into the place x could be in the different equation: -4((13 - 5y)/2) - 3y = 9 -26 + 10y - 3y = 9 7y = 35 y=5 Now pass back and locate x using the 1st equation: (13 - 5y)/2 = x = (13 - 25)/2 = x = -6 comparable element with the final one. Your x and y could be: 4) try this one with comparable approach and verify your solutions: x = -sixty 8, y = 39. desire this helps. I additionally desire i've got not made any blunders!

Answer 8

Elimination as it says we have to eliminate one variable from these two linear equations by applying simple math. In order to do so, we make co-efficient of one variable same in both equations. If these same co-efficient are of same sign we subtract one equation from other, if different sign we add them together
5x-4y=14
3y+3x=3
Re arranging
5x-4y=14
3x+3y=3
multiply first equation by 3 and second by 4
15x-12y=42
12x+12y=12
adding both of the above equations
27x=54 {we eliminated "y" here}
x=54/27=2
plug in X's value in any of the equation
3y+3*2=3
3y+6=3
3y=3-6
3y=-3
y=-3/3=-1
so
x=2
y=-1

Answer 9

You add the two equations together. Add the X's together, the Y's together, and the constants together. Make sure either the X's or the Y's equals 0. To do this you might have to multiply one equation by a number (possibly a negative number). If you don't do this it is substitution instead of elimination.

Answer 10

To use elimination, you will need to make the coefficient of y (or x if you want) same in boyh equation.
So multiply first one by 3, second one by 4
( this 3 and 4 are coefficient of y in the equation but in reversed order, by doing the multiplication we are making coefficient equal in both equation)
so

15x-12y=42
12x+12y=12

now add both equation to eliminate y, you will get the value of x.